Hyperbola Hidden within a Hyperboloid

A quadric surface is a surface with the equation \(Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0\). Depending on which coefficients are negative, positive, and zero, you can get ellipses, paraboloids, cylinders, cones, and hyperboloids out of this equation. What is interesting about quadric surfaces is that if you take a cross section of the surface, you can get a conic section (ellipse, parabola, or hyperbola) depending on how it's cut. The equation of the cross section is \(Ax^2+By^2+Cxy+Dx+Ey+F=0\). The way you find the equation of the cross section is a bit complicated. First, the equation of the quadric surface has to be simplified. If there is any \(G, H\), or \(I\), you must complete the square with the corresponding \(A, B,\) or \(C\). For any \(D, E\), or \(F\), you must replace \(x,y\),and \(z\) with the parameterized versions and use a \(\theta\) or \(\phi\) that will cancel out the \(xy, xz,\) and \(yz\). After that, put the equation of the plane in the form \(Ax + By + Cz = D\) and solve for one of the variables. Then, replace the variable in the simplified quadric surface equation and now you have the equation of a conic section. If there is any \(C, D\), or \(E\) in the conic section equation, do the same thing as before to get rid of them. After that, you have a recognizable formula for the cross section of the quadric surface.
 

The quadric surface I chose was a hyperboloid of one sheet with the equation \(\frac{x^2}{2^2} + \frac{y^2}{1^2}-\frac{z^2}{.5^2}=1\). This hyperboloid is twice as long in the \(x\) direction and half as tall in the \(z\) direction compared to a standard hyperboloid of one sheet. The intersecting plane is \(x+y+z=0\) which is a plane perpendicular to the vector [1,1,1]. The way my plane intersects the hyperboloid creates a hyperbola. Solving for \(z\) in the plane equation gets me \(z=-x-y\). Plugging that into the hyperboloid equation gets me \(\frac{x^2}{2^2} + \frac{y^2}{1^2}-\frac{(-x-y)^2}{.5^2}=1\). Simplifying gets \(\frac{-15x^2}{4} - 3y^2-8xy=1\). There is no \(D\) or \(E\) in this equation, so we just have to replace \(x\) and \(y\) with their parameterizations which are \(x=x\cos\theta-y\sin\theta\) and \(y=x\sin\theta+y\cos\theta\). The \(\theta\) needed for this simplification is \(\frac{1}{2}*\cot{\frac{-8}{-3.75+3}}\). I have no clue how to represent that in a better way, so from now on, we're gonna use decimal values rounded to four places. After plugging \(\theta\) in and simplifying, the equation we end up with is \(-7.3925x^2+.6425y^2=1\). This is the equation of a thin hyperbola with the vertices close to each other and on the y axis.

  The reason I chose this quadric surface and intersection is because I think that hyperbolas are the least utilized conic section. Sure, circles, ellipses, and parabolas have more real world practicality, but hyperbolas can be used to locate a point given its distance to three other points. A circle can't do that. As for the quadric surface, a hyperboloid is just a hyperbola rotated about its axis. So now, I get to demonstrate how to find the cross section of a quadric surface using not only a hyperbola, but also what is basically a hyperbola in 3D. I chose the coefficients for the hyperboloid and the intersecting plane to make finding the conic section a little more challenging. A standard hyperboloid and a level plane would be able to demonstrate how conic sections are found, but adding coefficients brings in a few more obstacles to tackle.